The energy subtraction method which uses the transmission intensity of a radiation source containing plural energy levels or energy bands is a conventional method of determining the quantity of specific substances in a physical object.
When the object being analyzed is composed of two different substances, the quantity of a specific constituent substance can be determined as follows. By way of example, an object 80 composed of substances identified simply as substance A and substance C as shown in FIG. 12 is to be analyzed.
Because the energy subtraction method is used, radiation 4 comprising two different energy levels E1 and E2 is irradiated to the object 80 being analyzed, and the corresponding transmission intensities values I.sub.1 and I.sub.2 are measured at predetermined points.
The transmission intensities I.sub.X1 and I.sub.X2 of the radiation 4 comprising two different energies E1 and E2 passing through the object 80 at point X in FIG. 12 can be expressed by equations 1 and 2 below. EQU I.sub.X1 =I.sub.01 exp(-.mu..sub.A1 .rho..sub.A T.sub.A -.mu..sub.C1 .rho..sub.C T.sub.C ') [1] EQU I.sub.X2 =I.sub.02 exp(-.mu..sub.A2 .rho..sub.A T.sub.A -.mu..sub.C2 .rho..sub.C T.sub.C ') [2]
The transmission intensities I.sub.Y1 and I.sub.Y2 after passing through the object 80 at point Y in FIG. 12 can be expressed by equations 3 and 4 below. EQU I.sub.Y1 =I.sub.01 exp(-.mu..sub.C1 .rho..sub.C T.sub.C) [3] EQU I.sub.Y2 =I.sub.02 exp(-.mu..sub.C2 .rho..sub.C T.sub.C) [4]
In each of the equations above, .mu..sub.A1, .mu..sub.A2, .mu..sub.C1, and .mu..sub.C2 are the mass attenuation coefficients of substances A and C at energy levels E1 and E2; I.sub.01 and I.sub.02 are the radiation intensities of the energy levels E1 and E2 irradiated to the object 80, respectively; .rho..sub.A and .rho..sub.C are the density of substances A and C, respectively; T.sub.A and T.sub.C are the thicknesses of substances A and C, respectively. T.sub.C ' is the thickness of substance C at X as defined by equation [5]. EQU T.sub.C '=T.sub.C -T.sub.A [ 5]
The density of substance A is obtained based on the energy subtraction calculation equation [6], which eliminates the effect of T.sub.C. EQU S=R.sub.C * In(I.sub.2 /I.sub.02)-ln(I.sub.1 /I.sub.01) [6]
where R.sub.C =.mu..sub.C1 /.mu..sub.C2.
Because the transmission intensities at point X are expressed by equations 1 and 2, the result of equation 6 is as shown by equation 7. ##EQU1##
The density m.sub.A per unit area of substance A can thus be obtained by equation 8. ##EQU2##
Because the transmission intensities at point Y are expressed by equations [3] and [4], calculating equation [6] for the value of S results in 0 (zero). It is therefore possible to identify the presence of substance A at each point of the object 80 using the subtraction calculation shown in equation [6], and the density can be quantified. By processing the resulting data, it is also possible to obtain an image showing the density of substance A relative to the other constituent substances.
When a third substance B 12 is also present in the object 10 being analyzed as shown in FIG. 1, it is now necessary obtain the density m.sub.B per unit area of this substance B 12 by calculating equations similar as those [7] and [8].
When the object being analyzed is composed of three types of substances in a layered construction whereby the radiation passes through all three substances, it is necessary to use a radiation source with three energy bands.
The process whereby an object 20 comprising three different substances A 21, B 22 and C 23, as shown in FIG. 2, is described next.
Because the energy subtraction method is used, a radiation 4 comprising three different energy bands E1, E2, and E3 is irradiated to the object 20 being analyzed.
The transmission intensities values I.sub.aa1, I.sub.aa2, and I.sub.aa3 of the radiation 4 comprising three different energy bands E1, E2, and E3 passing through the object 80 at point aa in FIG. 2 can be expressed by equations 9, 10, and 11 below. EQU I.sub.aa1 =I.sub.01 exp(-.mu..sub.A1 .rho..sub.A T.sub.A -.mu..sub.B1 .rho..sub.B T.sub.B -.mu..sub.C1 .rho..sub.C T.sub.C ') [9] EQU I.sub.aa2 =I.sub.02 exp(-.mu..sub.A2 .rho..sub.A T.sub.A -.mu..sub.B2 .rho..sub.B T.sub.B -.mu..sub.C2 .rho..sub.C T.sub.C ') [10] EQU I.sub.aa3 =I.sub.03 exp(-.mu..sub.A3 .rho..sub.A T.sub.A -.mu..sub.B3 .rho..sub.B T.sub.B -.mu..sub.C3 .rho..sub.C T.sub.C ') [11]
The transmission intensities I.sub.cc1, I.sub.cc2 and I.sub.cc3 after passing through the object at point cc in FIG. 2 can be expressed by equations 12, 13 and 14 below. EQU I.sub.cc1 =I.sub.01 exp(-.mu..sub.C1 .rho..sub.C T.sub.C) [12] EQU I.sub.cc2 =I.sub.02 exp(-.mu..sub.C2 .rho..sub.C T.sub.C) [13] EQU I.sub.cc3 =I.sub.03 exp(-.mu..sub.C3 .rho..sub.C T.sub.C) [14]
The transmission intensities I.sub.bb1, I.sub.bb2 and I.sub.bb3 after passing through the object at point bb in FIG. 2 can be expressed by equations 15, 16 and 17 below. EQU I.sub.bb1 =I.sub.01 exp(-.mu..sub.B1 .rho..sub.B T.sub.B -.mu..sub.C1 .rho..sub.C T.sub.C ") [15] EQU I.sub.bb2 =I.sub.02 exp(-.mu..sub.B2 .rho..sub.B T.sub.B -.mu..sub.C2 .rho..sub.C T.sub.C ") [16] EQU I.sub.bb3 =I.sub.03 exp(-.mu..sub.B3 .rho..sub.B T.sub.B -.mu..sub.C3 .rho..sub.C T.sub.C ") [17]
In each of the equations 9-17 above, .mu..sub.A1, .mu..sub.A2, .mu..sub.A3, .mu..sub.B1, .mu..sub.B2, .mu..sub.B3, .mu..sub.C1, .mu..sub.C2 and .mu..sub.C3 are the mass attenuation coefficients of substances A, B, and C at energy bands E1, E2 and E3, respectively; I.sub.01, I.sub.02 and I.sub.03 are the radiation intensities of the energy bands E1, E2, E3, respectively, irradiated to the object; .rho..sub.A, .rho..sub.B and .rho..sub.C are the densities of substances A, B and C, respectively; and T.sub.A, T.sub.B and T.sub.C are the thickness of substances A, B, and C, respectively. T.sub.C ' and T.sub.C " are defined by equations 18 and 19 below. EQU T.sub.C '=T.sub.C -T.sub.A -T.sub.B [ 18] EQU T.sub.C "=T.sub.C -T.sub.B [ 19]
The logarithm of both sides of equations 9-11 is obtained and expressed as follows. EQU L.sub.aa1 =-.mu..sub.A1 .rho..sub.A T.sub.A -.mu..sub.B1 .rho..sub.B T.sub.B -.mu..sub.C1 .rho..sub.C T.sub.C ' [20] EQU L.sub.aa2 =-.mu..sub.A2 .rho..sub.A T.sub.A -.mu..sub.B2 .rho..sub.B T.sub.B -.mu..sub.C2 .rho..sub.C T.sub.C ' [21] EQU L.sub.aa3 =-.mu..sub.A3 .rho..sub.A T.sub.A -.mu..sub.B3 .rho..sub.B T.sub.B -.mu..sub.C3 .rho..sub.C T.sub.C ' [22]
By eliminating T.sub.B and T.sub.C ' from these three equations and solving for T.sub.A, equation 23 is obtained. EQU m.sub.A =.rho..sub.A T.sub.A =D.sub.A /D [23]
where D.sub.A and D are defined as EQU D.sub.A =-.mu..sub.A1 .mu..sub.B2 .mu..sub.C3 -.mu..sub.B1 .mu..sub.C2 .mu..sub.A3 -.mu..sub.C1 .mu..sub.A2 .mu..sub.C3 EQU +.mu..sub.C1 .mu..sub.B2 .mu..sub.A3 +.mu..sub.A1 .mu..sub.C2 .mu..sub.B3 +.mu..sub.B1 .mu..sub.A2 .mu..sub.C3 [ 24] EQU D=S.sub.1 .mu..sub.B2 .mu..sub.C3 +.mu..sub.B1 .mu..sub.C2 S.sub.3 +.mu..sub.C1 S.sub.2 .mu..sub.C3 EQU -.mu..sub.C1 .mu..sub.B2 S.sub.3 -S.sub.1 .mu..sub.C2 .mu..sub.B3 -.mu..sub.B1 S.sub.2 .mu..sub.C3 [ 25]
Substance A 21 is not in the radiation transmission path at points cc and bb, and the transmission intensity of the radiation is expressed by equations [12]-[17]. Applying these values in equation [23], m.sub.A is 0 at points cc and bb.
The densities m.sub.B and m.sub.C per unit area of substances B 22 and C 23, respectively, are similarly obtained from equations [20]-[22].
The above method has thus been used to determine the quantities of specific substances in the object.
As described above, it is necessary to use a radiation source with as many energy bands as there are constituent substances in the object being analyzed. When X-ray radiation is used, energy separation using a k-edge filter is performed simultaneously to irradiation of an X-ray source with plural energy bands. Alternatively, the X-ray tube voltage is continuously switched to generate a series of X-ray beams of different energy bands.
When a k-edge filter is used to form plural energy bands, absorption by the k-edge filter reduces the photon number of each energy band, and the transmission image of the analyzed sample cannot be precisely measured. When the tube voltage is changed, excessive time is required for the measurement, and movement by the subject creates problems. In addition, the equations used in this conventional method must be varied for each substance analyzed. As a result, simultaneous quantification of plural substances in a single object is not possible in a single calculation process.